The problem asks us to prove the following trigonometric identity: $\sqrt{\frac{1 + \cos A}{1 - \cos A}} = \csc A + \cot A$

1. Problem Description

The problem asks us to prove the following trigonometric identity:

$\sqrt{\frac{1 + \cos A}{1 - \cos A}} = \csc A + \cot A$

2. Solution Steps

We will start with the left-hand side (LHS) of the equation and try to manipulate it to match the right-hand side (RHS).

LHS: $\sqrt{\frac{1 + \cos A}{1 - \cos A}}$

Multiply the numerator and denominator inside the square root by $(1 + \cos A)$:

$\sqrt{\frac{1 + \cos A}{1 - \cos A} \cdot \frac{1 + \cos A}{1 + \cos A}} = \sqrt{\frac{(1 + \cos A)^2}{(1 - \cos A)(1 + \cos A)}}$

Using the difference of squares formula, $(a - b)(a + b) = a^2 - b^2$, we have:

$\sqrt{\frac{(1 + \cos A)^2}{1 - \cos^2 A}}$

Using the Pythagorean trigonometric identity, $\sin^2 A + \cos^2 A = 1$, we can rewrite $1 - \cos^2 A$ as $\sin^2 A$:

$\sqrt{\frac{(1 + \cos A)^2}{\sin^2 A}}$

Now, we can take the square root of both the numerator and the denominator:

$\frac{1 + \cos A}{\sin A}$

Split the fraction into two parts:

$\frac{1}{\sin A} + \frac{\cos A}{\sin A}$

Recall the definitions of cosecant and cotangent:

$\csc A = \frac{1}{\sin A}$

$\cot A = \frac{\cos A}{\sin A}$

Therefore, we have:

$\csc A + \cot A$

Which is equal to the RHS.

3. Final Answer

$\sqrt{\frac{1 + \cos A}{1 - \cos A}} = \csc A + \cot A$

The given identity is proven.